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Re: variation on a familiar calculus problem
Posted:
May 8, 2012 10:37 PM
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If you set the condition that one of the sides must be perpendicular to the wall then that handles all odd N sided polygons. In Sandy's first problem (the rectangle) there were three sides but with the condition that it be rectangular (two sides perpendicular to the wall and one side parallel). If the condition was simply that one side be perpendicular to the wall then the answer would have been half of a regular pentagon with one side bisected by the wall and one vertice on the wall.
Bob Hansen
On May 8, 2012, at 11:02 AM, Joe Niederberger <niederberger@comcast.net> wrote:
>> More to what is going on here...
>
> Yeah, that's pretty good I think. And it does extend to odd-sided, as well as smaller divisions for even sided, but then you need two walls as I mentioned before. For a triangle, the "cost-free" walls have an angle
> of 2pi/3 (120 deg.) For a pentagon, 2pi/5. For a hexagon, you can choose to divide it in half, or thirds, etc. (and for the square, of course, you could also divide that into fourths...) I could be mistaken, only worked this out without paper, but your proof seems to solidify the thinking. Nice job!
>
> Joe N
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